Pano po naging sagot sa problem na integral of dx all over x(1+x^2) e 1/2 ln x^2 over 1+x^2 +c

hindi po malinaw kung ano talaga ang equation. You can use symbols of grouping (brackets) para mas madali maintindihan.

patulong nga po. 3x^4+4x^3+16x^2+20x+9 over (x+20)(x^2+3)^2 pasagot nga po yan hirap po kase

Ano ang tanong dito? Mahirap manghula kuna ano ang ibig mong sabihin.

Binasa ko uli ang post mo at naintindihan ko na. You ask why $\displaystyle \int \dfrac{dx}{x(1 + x^2)} = \dfrac{1}{2} \, \ln \left( \dfrac{x^2}{1 + x^2} \right) + C$

am I right? If so, here is the detail:

$\dfrac{1}{x(1 + x^2)} = \dfrac{A}{x} + \dfrac{Bx + C}{1 + x^2}$

$1 = A(1 + x^2) + Bx^2 + Cx$

When x = 0, A = 1 Equate x_{2}: 0 = A + B, B = -1 Equate x: C = 0

Thus, $\displaystyle \int \dfrac{dx}{x(1 + x^2)} = \int \left( \dfrac{1}{x} - \dfrac{x}{1 + x^2} \right) \, dx$

$\displaystyle \int \dfrac{dx}{x(1 + x^2)} = \int \dfrac{dd}{x} - \dfrac{1}{2} \int \dfrac{2x \, dx}{1 + x^2}$

$\displaystyle \int \dfrac{dx}{x(1 + x^2)} = \ln x - \dfrac{1}{2} \ln (1 + x^2) + C$

$\displaystyle \int \dfrac{dx}{x(1 + x^2)} = \dfrac{1}{2} \left[ 2\ln x - \ln (1 + x^2) \right] + C$

$\displaystyle \int \dfrac{dx}{x(1 + x^2)} = \dfrac{1}{2} \left[ \ln x^2 - \ln (1 + x^2) \right] + C$

$\displaystyle \int \dfrac{dx}{x(1 + x^2)} = \dfrac{1}{2} \, \ln \left( \dfrac{x^2}{1 + x^2} \right) + C$

as we expect it to be.

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hindi po malinaw kung ano talaga ang equation. You can use symbols of grouping (brackets) para mas madali maintindihan.

patulong nga po. 3x^4+4x^3+16x^2+20x+9 over (x+20)(x^2+3)^2 pasagot nga po yan hirap po kase

Ano ang tanong dito? Mahirap manghula kuna ano ang ibig mong sabihin.

Binasa ko uli ang post mo at naintindihan ko na. You ask why

$\displaystyle \int \dfrac{dx}{x(1 + x^2)} = \dfrac{1}{2} \, \ln \left( \dfrac{x^2}{1 + x^2} \right) + C$

am I right? If so, here is the detail:

$\dfrac{1}{x(1 + x^2)} = \dfrac{A}{x} + \dfrac{Bx + C}{1 + x^2}$

$1 = A(1 + x^2) + Bx^2 + Cx$

When x = 0, A = 1

Equate x

_{2}: 0 = A + B, B = -1Equate x: C = 0

Thus,

$\displaystyle \int \dfrac{dx}{x(1 + x^2)} = \int \left( \dfrac{1}{x} - \dfrac{x}{1 + x^2} \right) \, dx$

$\displaystyle \int \dfrac{dx}{x(1 + x^2)} = \int \dfrac{dd}{x} - \dfrac{1}{2} \int \dfrac{2x \, dx}{1 + x^2}$

$\displaystyle \int \dfrac{dx}{x(1 + x^2)} = \ln x - \dfrac{1}{2} \ln (1 + x^2) + C$

$\displaystyle \int \dfrac{dx}{x(1 + x^2)} = \dfrac{1}{2} \left[ 2\ln x - \ln (1 + x^2) \right] + C$

$\displaystyle \int \dfrac{dx}{x(1 + x^2)} = \dfrac{1}{2} \left[ \ln x^2 - \ln (1 + x^2) \right] + C$

$\displaystyle \int \dfrac{dx}{x(1 + x^2)} = \dfrac{1}{2} \, \ln \left( \dfrac{x^2}{1 + x^2} \right) + C$

as we expect it to be.

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